Integration of Hyperbolic Functions
Solved Problems
Example 9.
Evaluate the integral \[\int {{e^{-x}}\sinh 2xdx}.\]
Solution.
The hyperbolic sine function is defined as
\[\sinh x = \frac{{{e^x} - {e^{ - x}}}}{2}.\]
Hence the integral is given by
\[\int {{e^{ - x}}\sinh 2xdx} = \int {{e^{ - x}}\left( {\frac{{{e^{2x}} - {e^{ - 2x}}}}{2}} \right)dx} = \frac{1}{2}\int {\left( {{e^x} - {e^{ - 3x}}} \right)dx} = \frac{1}{2}\left[ {{e^x} - \frac{{{e^{ - 3x}}}}{{\left( { - 3} \right)}}} \right] + C = \frac{{{e^x}}}{2} + \frac{{{e^{ - 3x}}}}{6} + C.\]
Example 10.
Evaluate the integral \[\int {\frac{{dx}}{{\sinh x}}}.\]
Solution.
Using the identity
\[{\cosh ^2}x - {\sinh ^2}x = 1,\]
we rewrite the integral as follows:
\[I = \int {\frac{{dx}}{{\sinh x}}} = \int {\frac{{\sinh x}}{{{{\sinh }^2}x}}dx} = \int {\frac{{\sinh x}}{{{{\cosh }^2}x - 1}}dx} .\]
Make the substitution
\[u = \cosh x,\;\; \Rightarrow du = \sinh xdx.\]
Hence
\[I = \int {\frac{{du}}{{{u^2} - 1}}} = \frac{1}{2}\ln \left| {\frac{{u - 1}}{{u + 1}}} \right| + C = \frac{1}{2}\ln \left| {\frac{{\cosh x - 1}}{{\cosh x + 1}}} \right| + C = \ln \sqrt {\left| {\frac{{\cosh x - 1}}{{\cosh x + 1}}} \right|} + C.\]
We can simplify the expression under the square root. Since
\[\cosh x - 1 = {\cosh ^2}\frac{x}{2} + {\sinh ^2}\frac{x}{2} - \left( {{{\cosh }^2}\frac{x}{2} - {{\sinh }^2}\frac{x}{2}} \right) = \cancel{{\cosh ^2}\frac{x}{2}} + {\sinh ^2}\frac{x}{2} - \cancel{{\cosh ^2}\frac{x}{2}} + {\sinh ^2}\frac{x}{2} = 2\,{\sinh ^2}\frac{x}{2},\]
\[\cosh x + 1 = {\cosh ^2}\frac{x}{2} + \cancel{{\sinh ^2}\frac{x}{2}} + {\cosh ^2}\frac{x}{2} - \cancel{{\sinh ^2}\frac{x}{2}} = 2\,{\cosh ^2}\frac{x}{2},\]
we obtain
\[\frac{{1 - \cosh x}}{{1 + \cosh x}} = \frac{{\cancel{2}{{\sinh }^2}\frac{x}{2}}}{{\cancel{2}{{\cosh }^2}\frac{x}{2}}} = {\tanh ^2}\frac{x}{2}.\]
The integral is expressed in the form
\[I = \ln \sqrt {\left| {\frac{{\cosh x - 1}}{{\cosh x + 1}}} \right|} + C = \ln \sqrt {\left| {{{\tanh }^2}\frac{x}{2}} \right|} + C = \ln \left| {\tanh \frac{x}{2}} \right| + C.\]
Example 11.
Find the integral \[\int {\frac{{dx}}{{1 + \cosh x}}}.\]
Solution.
By definition of the hyperbolic cosine,
\[\cosh x = {\frac{{{e^x} + {e^{ - x}}}}{2}}.\]
Hence, the integral is equal
\[\int {\frac{{dx}}{{1 + \cosh x}}} = \int {\frac{{dx}}{{1 + \frac{{{e^x} + {e^{ - x}}}}{2}}}} = \int {\frac{{2dx}}{{2 + {e^x} + {e^{ - x}}}}} = 2\int {\frac{{{e^x}dx}}{{2{e^x} + {e^{2x}} + 1}}} = 2\int {\frac{{d\left( {{e^x} + 1} \right)}}{{{{\left( {{e^x} + 1} \right)}^2}}}} = - \frac{2}{{{e^x} + 1}} + C.\]
Example 12.
Find the integral \[\int {\frac{{dx}}{{\sinh x + 2\cosh x}}}.\]
Solution.
By definition,
\[\sinh x = \frac{{{e^x} - {e^{ - x}}}}{2} \;\;\text{and}\;\; \cosh x = \frac{{{e^x} + {e^{ - x}}}}{2}.\]
Hence,
\[I = \int {\frac{{dx}}{{\sinh x + 2\cosh x}}} = \int {\frac{{dx}}{{\frac{{{e^x} - {e^{ - x}}}}{2} + 2 \cdot \frac{{{e^x} + {e^{ - x}}}}{2}}}} = \int {\frac{{2dx}}{{{e^x} - {e^{ - x}} + 2{e^x} + 2{e^{ - x}}}}} = 2\int {\frac{{dx}}{{3{e^x} + {e^{ - x}}}}} = 2\int {\frac{{{e^x}dx}}{{3{e^{2x}} + 1}}} .\]
We make the substitution: \(u = {e^x},\) \(du = {e^x}dx\) and calculate the initial integral:
\[I = 2\int {\frac{{{e^x}dx}}{{3{e^{2x}} + 1}}} = 2\int {\frac{{du}}{{3{u^2} + 1}}} = \frac{2}{3}\int {\frac{{du}}{{{u^2} + \frac{1}{3}}}} = \frac{2}{3}\int {\frac{{du}}{{{u^2} + {{\left( {\frac{1}{{\sqrt 3 }}} \right)}^2}}}} = \frac{2}{3} \cdot \frac{{\sqrt 3 }}{1}\arctan \frac{u}{{\frac{1}{{\sqrt 3 }}}} + C = \frac{2}{{\sqrt 3 }}\arctan \left( {\sqrt 3 u} \right) + C = \frac{2}{{\sqrt 3 }}\arctan \left( {\sqrt 3 {e^x}} \right) + C.\]
Example 13.
Find the integral \[\int {\frac{{dx}}{{\sinh x - \cosh x}}}.\]
Solution.
Since
\[\sinh x = \frac{{{e^x} - {e^{ - x}}}}{2},\;\;\cosh x = \frac{{{e^x} + {e^{ - x}}}}{2},\]
the integral is expressed in the form
\[\int {\frac{{dx}}{{\sinh x - \cosh x}}} = \int {\frac{{dx}}{{\frac{{{e^x} - {e^{ - x}}}}{2} - \frac{{{e^x} + {e^{ - x}}}}{2}}}} = \int {\frac{{2dx}}{{\cancel{e^x} - {e^{ - x}} - \cancel{e^x} - {e^{ - x}}}}} = \int {\frac{{\cancel{2}dx}}{{ - \cancel{2}{e^{ - x}}}}} = - \int {{e^x}dx} = - {e^x} + C.\]
Example 14.
Find the integral \[\int {\frac{{dx}}{{3\sinh x - 5\cosh x}}}.\]
Solution.
We substitute the definitions of the hyperbolic sine and cosine functions in the integrand:
\[\sinh x = \frac{{{e^x} - {e^{ - x}}}}{2},\;\;\cosh x = \frac{{{e^x} + {e^{ - x}}}}{2}.\]
This yields:
\[I = \int {\frac{{dx}}{{3\sinh x - 5\cosh x}}} = \int {\frac{{dx}}{{3 \cdot \frac{{{e^x} - {e^{ - x}}}}{2} - 5 \cdot \frac{{{e^x} + {e^{ - x}}}}{2}}}} = \int {\frac{{2dx}}{{3{e^x} - 3{e^{ - x}} - 5{e^x} - 5{e^{ - x}}}}} = \int {\frac{{2dx}}{{ - 2{e^x} - 8{e^{ - x}}}}} = - \int {\frac{{dx}}{{{e^x} + 4{e^{ - x}}}}} = - \int {\frac{{dx}}{{{e^x} + 4{e^{ - x}}}}} .\]
Next, we multiply the numerator and denominator by \({e^x}:\)
\[I = - \int {\frac{{dx}}{{{e^x} + 4{e^{ - x}}}}} = - \int {\frac{{{e^x}dx}}{{{e^{2x}} + 4}}} ,\]
and make the substitution
\[u = {e^x},\;\; \Rightarrow du = {e^x}dx.\]
We obtain a basic integral:
\[I = - \int {\frac{{{e^x}dx}}{{{e^{2x}} + 4}}} = - \int {\frac{{du}}{{{u^2} + {2^2}}}} = - \frac{1}{2}\arctan \frac{u}{2} + C = - \frac{1}{2}\arctan \frac{{{e^x}}}{2} + C.\]
Example 15.
Evaluate the integral \[\int {\sinh x\cosh \left( { - x} \right)dx}.\]
Solution.
The hyperbolic cosine function is even:
\[\cosh \left( { - x} \right) = \cosh x.\]
Hence the integral is given by
\[\int {\sinh x\cosh \left( { - x} \right)dx} = \int {\sinh x\cosh xdx} = \int {\frac{{\sinh 2x}}{2}dx} = \frac{1}{2}\int {\sinh 2xdx} = \frac{1}{2} \cdot \frac{{\cosh 2x}}{2} + C = \frac{{\cosh 2x}}{4} + C.\]
Example 16.
Calculate the integral \[\int {\sinh 2x\cosh 3xdx}.\]
Solution.
Applying the formulas
\[\sinh x = \frac{{{e^x} - {e^{ - x}}}}{2} \;\;\text{and}\;\; \cosh x = \frac{{{e^x} + {e^{ - x}}}}{2},\]
we get
\[\int {\sinh 2x\cosh 3xdx} = \int {\frac{{{e^{2x}} - {e^{ - 2x}}}}{2} \cdot \frac{{{e^{3x}} + {e^{ - 3x}}}}{2}dx} = \frac{1}{4}\int {\left( {{e^{5x}} - {e^x} + {e^{ - x}} - {e^{ - 5x}}} \right)dx} = \frac{1}{4}\left( {\frac{{{e^{5x}}}}{5} - {e^x} - {e^{ - x}} + \frac{{{e^{ - 5x}}}}{5}} \right) + C = \frac{1}{{10}} \cdot \frac{{{e^{5x}} + {e^{ - 5x}}}}{2} - \frac{1}{2} \cdot \frac{{{e^x} + {e^{ - x}}}}{2} + C = \frac{{\cosh 5x}}{{10}} - \frac{{\cosh x}}{2} + C.\]
Example 17.
Find the integral \[\int {\tanh 2xdx}.\]
Solution.
Given that \(\tanh x = \frac{{\sinh x}}{{\cosh x}},\) we can write
\[I = \int {\tanh 2xdx} = \int {\frac{{\sinh 2x}}{{\cosh 2x}}dx} .\]
Make the substitution
\[u = \cosh 2x,\;\; \Rightarrow du = 2\sinh 2xdx,\;\; \Rightarrow \sinh 2xdx = \frac{{du}}{2}.\]
This yields:
\[I = \int {\frac{{\sinh 2x}}{{\cosh 2x}}dx} = \frac{1}{2}\int {\frac{{du}}{u}} = \frac{1}{2}\ln \left| u \right| + C = \ln \sqrt {\left| u \right|} + C = \ln \sqrt {\left| {\cosh 2x} \right|} + C.\]
Note that the hyperbolic cosine function is always positive. Hence,
\[I = \ln \sqrt {\cosh 2x} + C.\]
Example 18.
Find the integral \[\int {\coth \frac{x}{3} dx}.\]
Solution.
By definition,
\[\coth x = \frac{{\cosh x}}{{\sinh x}}.\]
Hence, we can write
\[I = \int {\coth \frac{x}{3}dx} = \int {\frac{{\cosh \frac{x}{3}}}{{\sinh \frac{x}{3}}}dx} .\]
Use the substitution
\[u = \sinh \frac{x}{3},\;\; \Rightarrow du = \frac{1}{3}\cosh \frac{x}{3}dx,\;\; \Rightarrow \cosh \frac{x}{3}dx = 3du.\]
Then the integral is given by
\[I = \int {\frac{{\cosh \frac{x}{3}}}{{\sinh \frac{x}{3}}}dx} = \int {\frac{{3du}}{u}} = 3\ln \left| u \right| + C = 3\ln \left| {\sinh \frac{x}{3}} \right| + C.\]
Example 19.
Evaluate the integral \[\int {\sinh x\cos xdx}.\]
Solution.
We use integration by parts. Let
\[u = \cos x,\;\;dv = \sinh xdx,\;\; \Rightarrow du = - \sin xdx,\;\; v = \int {\sinh xdx} = \cosh x.\]
Then
\[\int {\sinh x\cos xdx} = \cosh x\cos x - \int {\cosh x\left( { - \sin x} \right)dx} = \cosh x\cos x + \int {\cosh x\sin xdx}.\]
Apply integration by parts again to the latter integral. Let
\[u = \sin x,\;\; dv = \cosh xdx,\;\; \Rightarrow du = \cos xdx,\;\; v = \int {\cosh xdx} = \sinh x.\]
Then we have
\[\int {\sinh x\cos xdx} = \cosh x\cos x + \int {\cosh x\sin xdx} = \cosh x\cos x + \left( {\sin x\sinh x - \int {\sinh x\cos xdx} } \right).\]
Solving this equation for \(\int {\sinh x\cos xdx},\) we obtain the complete answer:
\[\int {\sinh x\cos xdx} = \frac{{\cosh x\cos x + \sinh x \sin x}}{2}.\]
